'When Are Relations Neither True Nor False?
in [Math]
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From: Newberry on 24 Mar 2010 12:15 On Mar 23, 4:55 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Newberry <newberr...(a)gmail.com> writes: > > It is meaningless only if ~(Ex)Px is necessarly true. If ~(Ex)Px is > > contingent then > > > (Ax)(Px -> Qx) > > > is merely neither true nor false. In arithmetic it would indeed be > > meaningless. > > Okay. I don't recall you mentioning this distinction previously (and > I may well forget it hence). > > Thus. the sentence > > ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) > > is meaningless, right? You don't have any idea what it means? > Despite the fact that the classical proof that sqrt(2) is irrational > proceeds thus: > > |- (E a,b)( a^2/b^2 = 2 ) -> (E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) > |- ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) > -------------------------------------------------------------------- > So, ~(E a,b)( a^2/b^2 = 2 ) > 1) It lloks like you assume (E a,b)( a^2/b^2 = 2 ) and derive a contradiction. 2) How do you derive (E a,b)( a^2/b^2 = 2 ) -> (E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) ?
From: Jesse F. Hughes on 24 Mar 2010 12:22 Newberry <newberryxy(a)gmail.com> writes: > On Mar 23, 4:55 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Newberry <newberr...(a)gmail.com> writes: >> > It is meaningless only if ~(Ex)Px is necessarly true. If ~(Ex)Px is >> > contingent then >> >> > (Ax)(Px -> Qx) >> >> > is merely neither true nor false. In arithmetic it would indeed be >> > meaningless. >> >> Okay. I don't recall you mentioning this distinction previously (and >> I may well forget it hence). >> >> Thus. the sentence >> >> ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) >> >> is meaningless, right? You don't have any idea what it means? >> Despite the fact that the classical proof that sqrt(2) is irrational >> proceeds thus: >> >> |- (E a,b)( a^2/b^2 = 2 ) -> (E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) >> |- ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) >> -------------------------------------------------------------------- >> So, ~(E a,b)( a^2/b^2 = 2 ) >> > 1) It lloks like you assume (E a,b)( a^2/b^2 = 2 ) and derive a > contradiction. No, I didn't. You see the proof. It's a simple case of modus tollens. > 2) How do you derive > (E a,b)( a^2/b^2 = 2 ) -> (E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) ? It follows from the theorem (A a,b)(E c,d)( a/b = c/d and gcd(c,d) = 1 ). That is, any rational number can be written as c/d where c and d are coprime integers. -- "Am I am [sic] misanthrope? I would say no, for honestly I never heard of this word until about 1994 or thereabouts on the Internet reading a post from someone who called someone a misanthrope." -- Archimedes Plutonium
From: Jesse F. Hughes on 24 Mar 2010 12:34 Newberry <newberryxy(a)gmail.com> writes: > On Mar 24, 3:58 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > wrote: >> Newberry says... >> >> >> >> >> >> >> >> >On Mar 23, 5:42=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >> >wrote: >> >> The problem with this talk about statements being "meaningless" is >> >> that often the only way that you *know* that ~(Ex)Px is necessarly true >> >> is by proving a pair of statements: >> >> >> 1. (Ax) Px -> Qx >> >> >> 2. (Ax) Px -> ~Qx >> >> >> together, these imply >> >> >> 3. ~(Ex) Px >> >> >> which you claim implies that 1&2 are meaningless. >> >> >Maybe it is like a proof by contradiction. You assume (Ex)Px and >> >derive ~(Ex)Px. >> >> >How do you know that there are no alternative proofs? >> >> I don't know that there are no alternative proofs, but I >> know that the proof using 1-3 works perfectly fine. I can't >> see any *advantage* to calling some statements meaningless, >> especially when you can derive true consequences from them. >> >> >> I think it makes more sense to say that every statement >> >> of arithmetic is meaningful. >> >> >No, system with gaps has better properties than a system without gaps. >> >> Better in what sense? > > It can solve the Liar paradox. Perhaps you can remind us how that goes? How do you represent the Liar's paradox in your logic? > Tarki's theorem does not apply to it. > >> >> >Plus >> >> > (x)((x = x + 1) -> (x = x + 2)) >> >> >does not look particularly meaningful to me. >> >> The classical meaning of an implication > > The classical "meaning of an implication is known to be paradixical. Paradoxical in what sense? >> "A implies B" >> is that it is *not* the case that A is true and B is false. >> So for A implies B to be true, there are three possibilities: >> >> 1. A is true and B is true. >> 2. A is false and B is true. >> 3. A is false and B is false. >> >> In the case x=x+1 -> x=x+2, both are false, so we are >> in case 3. There is nothing meaningless about this. It's >> not a particularly *interesting* observation. But who >> cares? > > Right on! No harm will be done to arithmetic if such sentences are not > provable. This particular sentence does not appear useful. The fact that I can prove ~(E a b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) is historically useful. I use it to prove that sqrt(2) is irrational. You really need to stop pretending that if you show us a single "useless" sentence of the form ~(E x)(Px & Qx), then all such sentences are useless (when ~(Ex)Px is true). We actually use some of these to prove interesting facts -- including ~(E x)Px. Moreover, you said something stronger than that it is not provable. You said that it's meaningless. Let me give you another example. I suppose you agree that the following is a true statement about sets: (Ax)( (Ay)(y in x -> s(y) in x) -> (x = {} or x is infinite) ). That is, any set closed under the successor operation is either infinite or empty. Suppose we could prove that a particular finite set is closed under the successor operation. That is, that (Ay)(y in x -> s(y) in x) and x is not infinite. We would conclude that x is empty, right? Except, we couldn't conclude that, since if x is empty, (Ay)(y in x -> s(y) in x) is neither true nor false. Oops! Evidently, the empty set is not closed under the successor operation (according to you), since (Ay)( y in {} -> s(y) in {} ) is neither true nor false. So my question is this: since the empty set is not closed under the successor operation, would you say that (Ax)( (Ay)(y in x -> s(y) in x) -> (x is infinite) ) is a true statement? (I hope not.) If not, why not? Normally, I'd think that the following reasoning is valid: If (A -> (B or C)) and B -> "A is not true", then A -> C. Am I wrong? -- Jesse F. Hughes "For a gentle introduction to set theory, see Bourbaki (1970)." -- Footnote from "Transgressing the Boundaries", Alan Sokal
From: Daryl McCullough on 24 Mar 2010 14:15 Newberry says... > >On Mar 24, 3:58=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> >No, system with gaps has better properties than a system without gaps. >> >> Better in what sense? > >It can solve the Liar paradox. Tarki's theorem does not apply to it. The Liar sentence is not expressible in any of the usual mathematical theories---PA, or ZFC. Why is it preferable to have a language in which the Liar sentence is expressible? -- Daryl McCullough Ithaca, NY
From: Newberry on 25 Mar 2010 00:17
On Mar 24, 11:15 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Newberry says... > > > > >On Mar 24, 3:58=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >wrote: > >> >No, system with gaps has better properties than a system without gaps.. > > >> Better in what sense? > > >It can solve the Liar paradox. Tarki's theorem does not apply to it. > > The Liar sentence is not expressible in any of the usual mathematical > theories---PA, or ZFC. Why is it preferable to have a language in which > the Liar sentence is expressible? If you take the position that there are truth value gaps then the Liar papradox is solvable in English. Tarski's theorem does not apply to formal systems with gaps. I think it is preferable. > -- > Daryl McCullough > Ithaca, NY |